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3 years ago

12

Which formula correctly expresses the property density?

1 answer:

Olenka [21]3 years ago

6 0

Answer:

Option A

D = m/v

Explanation:

Density is defined as mass per unit volume of an object. Therefore, D=m/v where m is the mass of the object and v is the volume

Therefore, option A is the right option

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A torque of 50 N*m acts on a wheel of moment of inertia 25 kg*m^2 for 4 s and then is removed.

motikmotik

You have to find the calculate<span> the circumference first then you can just multiply the diameter by π, which is about 3.142. That gives you the distance for each </span>revolution<span>. Then you can multiply by the </span>number of revolutions<span> per minute.

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3 years ago

An electric iron operating at 120 volts draws 10. amperes of current. how much heat energy

faltersainse [42]

The complete question is how much heat energy is delivered by the iron in 30 seconds.
Heat is given by IVt, Where I is the current in Amperes, V is the voltage, and t is the time in seconds,
Therefore;
Heat energy = 120 × 10 ×30
= 36000 joules or 36 kJ
Heat energy is measured in joules.

8 0

3 years ago

Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which wou

Mariulka [41]

Answer:

w = 1.976 rpm

Explanation:

For simulate the gravity we will use the centripetal aceleration $a_c$ , so:

$a_c = w^2r$

where w is the angular aceleration and r the radius.

We know by the question that:

r = 60.5m

$a_c$ = 2.6m/s2

So, Replacing the data, and solving for w, we get:

$2.6m/s = w^2(60.5m)$

W = 0.207 rad/s

Finally we change the angular velocity from rad/s to rpm as:

W = 0.207 rad/s = 0.207*60/(2 $\pi$ )= 1.976 rpm

3 0

3 years ago

A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m3 , undergoes a constant-pressure expansion at

Gnom [1K]

Answer:

Work: 4.0 kJ, heat: 4.25 kJ

Explanation:

For a gas transformation at constant pressure, the work done by the gas is given by

$W=p(V_f -V_i)$

where in this case we have:

$p = 2 bar = 2\cdot 10^5 Pa$ is the pressure

$V_i = 0.1 m^3$ is the initial volume

$V_f = 0.12 m^3$ is the final volume

Substituting,

$W=(2\cdot 10^5)(0.12-0.10)=4000 J = 4.0 kJ$

The 1st law of thermodynamics also states that

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the gas

Q is the heat absorbed by the gas

Here we know that

$\Delta U = +0.25 kJ$

Therefore we can re-arrange the equation to find the heat absorbed by the gas:

$Q=\Delta U + W = 0.25 kJ + 4.0 kJ = 4.25 kJ$

7 0

4 years ago

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

3 years ago